RowQ
The Vault
RowQ
The Vault
CBSE Class 11 Chemistry · 10 questions · 24 marks
Chemistry runs on counting particles you cannot see, and the mole is the bridge that makes that possible. In this chapter you learn to move confidently between mass, moles, volume and number of particles, and to use that skill to fix formulae, balance stoichiometry and find out which reactant runs out first.
A sample of an oxide of nitrogen weighs 4.6 g and contains 1.4 g of nitrogen. Its empirical formula is:
Answer
NO₂. Moles of N = 1.4 ÷ 14 = 0.10 mol. Mass of oxygen = 4.6 − 1.4 = 3.2 g, so moles of O = 3.2 ÷ 16 = 0.20 mol. The N : O ratio is 0.10 : 0.20 = 1 : 2, giving the empirical formula NO₂. NO would need equal moles, N₂O needs O half of N, and N₂O₅ needs a 2 : 5 ratio, so none of those fit.
How many oxygen atoms are present in 0.25 mol of Al₂(SO₄)₃?
Answer
1.81 × 10²⁴ oxygen atoms. Each formula unit of Al₂(SO₄)₃ contains 3 × 4 = 12 oxygen atoms, so 0.25 mol of the compound contains 0.25 × 12 = 3.0 mol of O atoms. Number of atoms = 3.0 × 6.022 × 10²³ = 1.81 × 10²⁴. The other options correspond to forgetting the factor 12 or using only one sulphate group.
What volume of 0.50 M NaOH solution must be diluted to prepare 250 mL of 0.20 M NaOH?
Answer
100 mL. On dilution the moles of solute stay constant, so M₁V₁ = M₂V₂: (0.50)(V₁) = (0.20)(250 mL). V₁ = 50 ÷ 0.50 = 100 mL. So 100 mL of the 0.50 M stock is taken and water is added up to the 250 mL mark.
The result of the calculation (3.24 g) ÷ (1.2 cm³), reported to the correct number of significant figures, is:
Answer
2.7 g/cm³. The arithmetic gives 3.24 ÷ 1.2 = 2.70 g/cm³, but in multiplication and division the answer must carry the same number of significant figures as the least precise value. Here 1.2 has only two significant figures, so the result is rounded to two significant figures: 2.7 g/cm³.
Assertion (A): The molality of an aqueous solution does not change when the solution is warmed from 25 °C to 60 °C, but its molarity does. Reason (R): Molality is defined using the mass of solvent, which is independent of temperature, whereas molarity uses the volume of the solution, which expands on heating.
Answer
Both A and R are true and R is the correct explanation of A. Molality is moles of solute per kilogram of solvent, and mass does not change with temperature, so molality is a temperature-independent concentration term. Molarity is moles of solute per litre of solution, and the solution expands on heating, so the same moles now occupy a larger volume and the molarity falls slightly. The reason therefore explains exactly why one quantity changes and the other does not.
Calculate the mass of sodium carbonate (Na₂CO₃) required to prepare 500 mL of a 0.15 M solution. (Molar mass = 106 g/mol)
Answer
Moles required = M × V(in L) = 0.15 × 0.500 = 0.075 mol. Mass = moles × molar mass = 0.075 × 106 = 7.95 g. So 7.95 g of Na₂CO₃ is weighed, dissolved in a little water, and made up to exactly 500 mL in a volumetric flask.
A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Its molar mass is 180 g/mol. Determine its empirical and molecular formulae.
Answer
Take 100 g of the compound, so the masses are 40.0 g C, 6.7 g H and 53.3 g O. Moles: C = 40.0 ÷ 12 = 3.33; H = 6.7 ÷ 1 = 6.7; O = 53.3 ÷ 16 = 3.33. Dividing each by the smallest (3.33) gives C : H : O = 1 : 2 : 1, so the empirical formula is CH₂O with empirical formula mass 12 + 2 + 16 = 30 g/mol. n = 180 ÷ 30 = 6, so the molecular formula is C₆H₁₂O₆.
In a laboratory, 12.0 g of magnesium is burnt in a sealed container holding 5.6 g of oxygen gas to form magnesium oxide, MgO. (i) Write the balanced equation. (ii) Identify the limiting reagent with working. (iii) Calculate the mass of MgO formed. (iv) Find the mass of the excess reactant left over. (v) If only 12.6 g of MgO is actually collected, calculate the percentage yield. (Atomic masses: Mg = 24, O = 16)
Answer
(i) The balanced equation is 2Mg + O₂ → 2MgO. (ii) Moles of Mg = 12.0 ÷ 24 = 0.50 mol. Moles of O₂ = 5.6 ÷ 32 = 0.175 mol. The equation requires Mg and O₂ in a 2 : 1 mole ratio, so 0.50 mol of Mg would need 0.25 mol of O₂, but only 0.175 mol is available. Oxygen is therefore the limiting reagent, and magnesium is in excess. (iii) All calculations are based on the limiting reagent. 0.175 mol of O₂ produces 2 × 0.175 = 0.35 mol of MgO. Molar mass of MgO = 24 + 16 = 40 g/mol, so mass of MgO = 0.35 × 40 = 14.0 g. (iv) Mg consumed = 2 × 0.175 = 0.35 mol, which is 0.35 × 24 = 8.4 g. Mg left over = 12.0 − 8.4 = 3.6 g (0.15 mol). Conservation of mass checks out: 12.0 + 5.6 = 17.6 g total, and 14.0 g of MgO plus 3.6 g of unreacted Mg is also 17.6 g. (v) Percentage yield = (actual ÷ theoretical) × 100 = (12.6 ÷ 14.0) × 100 = 90%. The missing 10% is typical of transfer losses, magnesium that stays unburnt inside the ribbon, or a little magnesium reacting with nitrogen present in the vessel to form Mg₃N₂ instead.
State the law of multiple proportions. Two oxides of a metal M contain 63.6% and 46.7% of M by mass respectively. Show that these data illustrate the law, and comment on how such data helped early chemists assign formulae.
Answer
The law of multiple proportions states that when two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element bear a simple whole-number ratio to one another. Oxide 1: in 100 g there is 63.6 g of M and 36.4 g of O. Mass of O per 1 g of M = 36.4 ÷ 63.6 = 0.572 g. Oxide 2: in 100 g there is 46.7 g of M and 53.3 g of O. Mass of O per 1 g of M = 53.3 ÷ 46.7 = 1.141 g. Ratio of oxygen masses for a fixed mass of M = 0.572 : 1.141 = 1 : 1.995, which is essentially 1 : 2, a simple whole-number ratio. The data therefore obey the law of multiple proportions. Such ratios were historically vital because chemists could not weigh individual atoms. A clean 1 : 2 ratio strongly suggested that if the first oxide were MO, the second must be MO₂, so combining-mass data plus these laws let early chemists build a consistent table of relative atomic masses and formulae long before atoms could be observed directly.
A water-testing technician prepares a standard solution to check the hardness of a village borewell. She dissolves 3.70 g of calcium chloride (CaCl₂, molar mass 111 g/mol) in water and makes the volume up to 2.00 L in a volumetric flask. She then withdraws 25.0 mL of this solution for a titration. Answer: (a) Calculate the molarity of the prepared solution. (b) How many moles of CaCl₂ are present in the 25.0 mL sample? (c) How many chloride ions are present in that 25.0 mL sample? (d) She realises the flask was filled to 2.10 L by mistake — state qualitatively how this affects the reported molarity and why care with volumetric glassware matters.
Answer
(a) Moles of CaCl₂ = 3.70 ÷ 111 = 0.0333 mol. Molarity = 0.0333 ÷ 2.00 = 0.0167 M (about 1.67 × 10⁻² M). (b) Moles in 25.0 mL = 0.0167 × 0.0250 = 4.17 × 10⁻⁴ mol of CaCl₂. (c) Each formula unit gives 2 Cl⁻ ions, so moles of Cl⁻ = 2 × 4.17 × 10⁻⁴ = 8.33 × 10⁻⁴ mol. Number of ions = 8.33 × 10⁻⁴ × 6.022 × 10²³ = 5.02 × 10²⁰ chloride ions. (d) With 2.10 L instead of 2.00 L the same moles are spread over a larger volume, so the true molarity is 0.0333 ÷ 2.10 = 0.0159 M, about 5% lower than reported. Every titration result calculated from the assumed 0.0167 M would then be systematically about 5% too high. This is why volumetric flasks are filled exactly to the calibration mark with the bottom of the meniscus on the line — a small volume error propagates directly into every downstream concentration.
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